A particle moves along the curve $y = (\sqrt{x})^3$ in the first quadrant in such a way that its distance from the origin increases at the rate of $11$ units per second. The value of $\frac{dx}{dt}$ when $x=3$ is?
In the given solution, they had taken the distance from the origin to a point on the curve (green segment) and used the distance formula from one point to another, but in my opinion that is not actually distance, that is displacement. Actual distance is the red segment from $(0, 0)$ to $(3, 3^{3/2}).$ Am I going in the right direction or do they mean displacement when they said distance?
The way I calculated $\frac{dx}{dt}$ gives an approximated and less value than the given solution which makes sense, like displacement $\le$ distance, so $x$ value is going to be lesser in distance than in displacement or equal when they cover the same magnitude $y$.
Please let me know if you would like to see my solution and the solution that I have doubt in.
P.S:- Thank you @Dave L. Renfro @DavidQuinn @ryang for the clarification.
Best Answer
Writing them out unambiguously, do you see that these three expressions have different meanings?
$\color\red{\textbf{distance from the origin}}$
= magnitude of position
= magnitude of displacement relative to the origin
$\boldsymbol{P_1P_2}$
= distance between $\boldsymbol{P_1}$ and $\boldsymbol{P_2}$
= magnitude of displacement of $P_2$ relative to $P_1$
length of arc $\boldsymbol{P_1P_2}$
= distance along the path between $\boldsymbol{P_1}$ and $\boldsymbol{P_2}$
= distance travelled over $\boldsymbol{[t_1,t_2]}$
So, the given solution, as described, is correct, and should, by the chain rule, finally give $$\dfrac{\mathrm dx}{\mathrm dt}\Bigg|_{x=3}=\dfrac{\mathrm d|\boldsymbol r|}{\mathrm dt}\Bigg|_{x=3}\div \dfrac{\mathrm d|\boldsymbol r|}{\mathrm dx}\Bigg|_{x=3}=4.$$
Related: Deriving a particle's speed from its distance function.